a) \(R_1nt(R_2//R_3)\)

\(R_1=\dfrac{U_1}{I_1}\Rightarrow U_1=0,4.14=5,6\left(V\right)\)

\(I_1=I_{AB}=0,4A\)

Có \(R_{AB}=R_1+R_{23}=14+\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{434}{19}\left(\Omega\right)\)

\(\Rightarrow U_1+U_{23}=U_{AB}=R_{AB}.I_{AB}=\dfrac{439}{19}.0,4=\dfrac{868}{95}\left(V\right)\)

\(\Rightarrow U_{23}=\dfrac{868}{95}-5,6=\dfrac{336}{95}\left(V\right)\)

\(\Rightarrow U_2=U_3=\dfrac{336}{95}\left(V\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{24}{95}\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{14}{95}\left(V\right)\)

b) \(U_{AB}=\dfrac{868}{95}\left(V\right)\)

\(U_{AC}=I_1.R_1=0,4.14=5,6\left(V\right)\)

\(U_{CB}=I_{23}.R_{23}=0,4.\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{336}{95}\left(V\right)\)

Vậy...

Ủa sao bạn Thịnh kì zậy, tl như nào mà bỏ sót 2 câu luôn kìa :>

Đặt \(t=cosx;t\in\left[-1;1\right]\)

Để hàm số có tập xác định R

\(\Leftrightarrow cosx^2-\left(2+m\right)cosx+2m\ge0;\forall x\)

\(\Leftrightarrow t^2-\left(2+m\right)t+2m\ge0\) với mọi \(t\in\left[-1;1\right]\)

Đặt \(f\left(t\right)=t^2-\left(2+m\right)t+2m\); \(I\left(\dfrac{2+m}{2};f\left(\dfrac{2+m}{2}\right)\right)\)

TH1: \(\dfrac{2+m}{2}< -1\) \(\Leftrightarrow m< -4\)

Để \(f\left(t\right)\ge0;\forall t\in\left[-1;1\right]\) \(\Leftrightarrow\)\(f\left(t\right)_{min}=f\left(-1\right)\ge0\) \(\Leftrightarrow3+3m\ge0\Leftrightarrow m\ge-1\)(ktm đk)

TH2: \(-1\le\dfrac{m+2}{2}\le1\)\(\Leftrightarrow-4\le m\le0\)

Để \(f\left(t\right)\ge0;\forall t\in\left[-1;1\right]\) \(\Leftrightarrow f\left(t\right)_{min}=f\left(\dfrac{2+m}{2}\right)\ge0\)\(\Leftrightarrow-m^2+4m-4\ge0\)\(\Leftrightarrow m=2\) (ktm đk)

TH3:\(\dfrac{m+2}{2}>1\) \(\Leftrightarrow m>0\)

Để \(f\left(t\right)\ge0;\forall t\in\left[-1;1\right]\)\(\Leftrightarrow f\left(t\right)_{min}=f\left(1\right)\ge0\)\(\Leftrightarrow m-1\ge0\Leftrightarrow m\ge1\)

Kết hợp cả ba TH \(\Rightarrow m\ge1\)

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Với \(n>0;n\in N:\dfrac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\dfrac{1}{\sqrt{n\left(n+4\right)}\left(\sqrt{n}+\sqrt{n+4}\right)}=\dfrac{\sqrt{n+4}-\sqrt{n}}{\sqrt{n\left(n+4\right)}\left(n+4-n\right)}=\dfrac{1}{4}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+4}}\right)\) (1)

Áp dụng (1) ta được:

 \(\dfrac{1}{1\sqrt{5}+5\sqrt{1}}+\dfrac{1}{5\sqrt{9}+9\sqrt{5}}+...+\dfrac{1}{2013\sqrt{2017}+2017\sqrt{2013}}\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}-\dfrac{1}{\sqrt{9}}+...+\dfrac{1}{\sqrt{2013}}-\dfrac{1}{\sqrt{2017}}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{\sqrt{2017}}\right)=\dfrac{\sqrt{2017}-1}{4\sqrt{2017}}=\dfrac{2017-\sqrt{2017}}{8068}\)

Ý A

Có \(p+n+e=115\)

\(\Leftrightarrow2p+n=115\) (1)

Lại có:\(\dfrac{n}{p}=\dfrac{9}{7}\Rightarrow7n-9p=0\) (2)

Từ (1) (2) \(\Rightarrow\left\{{}\begin{matrix}n+2p=115\\7n-9p=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}n=45\\p=35\end{matrix}\right.\) 

R là Brom

a,Pt \(\Leftrightarrow cosx-sinx=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=-\dfrac{\pi}{4}-arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) ,\(k\in Z\)

b) Pt \(\Leftrightarrow\dfrac{4}{5}cosx-\dfrac{3}{5}sinx=\dfrac{3}{5}\)

Đặt \(cosa=\dfrac{4}{5}\Rightarrow sina=\dfrac{3}{5}\)

Pttt:\(cosx.cosa-sina.sinx=\dfrac{3}{5}\)

\(\Leftrightarrow cos\left(x+a\right)=\dfrac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-a+arc.cos\left(\dfrac{3}{5}\right)+2k\pi\\x=-a-arc.cos\left(\dfrac{3}{5}\right)+2k\pi\end{matrix}\right.\)(\(k\in Z\))

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c) Pt\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}.sin3x=1\)

Đặt \(cosa=\dfrac{3}{5}\Rightarrow sina=\dfrac{4}{5}\)

Pttt:\(cos3x.cosa+sin3a.sina=1\)

\(\Leftrightarrow cos\left(3x-a\right)=1\)

\(\Leftrightarrow x=\dfrac{a}{3}+\dfrac{k2\pi}{3}\)(\(k\in Z\))

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